# Minimal polynomial for any power of Jordan block is same as the minimal polynomial of the Jordan block.

Let $$J$$ be the $$n \times n$$ Jordan block corresponding to the eigen value $$1$$. For any natural number $$r$$ is it true that the minimal polynomial for $$J^r$$ is $$(X-1)^n$$ ?

Another way to think about it to produce a cyclic vector of $$J^r$$. I can’t prove it. I need some help. Thanks.

• This is true for any non-zero eigenvalue, not just eigenvalue $1$. – user593746 Dec 6 '18 at 11:19
• Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+\sum_{k=1}^r {{r}\choose{k}} N^k$. You can show that $\sum_{k=1}^r {{r}\choose{k}} N^k$ is nilpotent of order $n$. – Eric Dec 6 '18 at 15:40
• The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$. – ancientmathematician Dec 6 '18 at 16:34
• My interest is in 0 characteristic. Thanks for your counter example in positive characteristic. – user371231 Dec 6 '18 at 18:46

Hint: write $$J=I+N$$ where $$N$$ is the shift matrix. $$N$$ is nilpotent with index $$n$$. Now expand $$J^r=(I+N)^r=...$$ and find out what is the smallest $$m$$ we need in order to $$(J^r-I)^m=0$$.
As $$r(J-I)=r(J^r-I)$$, so geometric multiplicity is $$1$$in both case are same and hence same minimal polynomial. Here $$r$$ means rank of matrix.
• It is not true. Take $3\times 3$ case: $$J-I=\begin{bmatrix}0 & 1& 0\\0 & 0 & 1\\0 & 0 & 0\end{bmatrix},\quad J^2-I=\begin{bmatrix}0 & 2& \color{red}{1}\\0 & 0 & 2\\0 & 0 & 0\end{bmatrix}.$$ In what sense are they equal? – A.Γ. Dec 6 '18 at 12:50
• Both has same rank as $2.$ – neelkanth Dec 6 '18 at 13:03